leetcode刷题笔记十一 盛最多水的容器 Scala版本
源地址：盛最多水的容器
问题描述：
Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container and n is at least 2.
The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
Example:
Input: [1,8,6,2,5,4,8,3,7]
Output: 49
简要思路分析：
首先想到，可以通过暴力法计算任意两条线与X轴构成的面积，通过变量记录其中的最大值，并返回。但是这种复杂度较高。
易知，最终面积是由连续的点构成的，我们可以视为最终面积是从起始点到终点构成的最初面积，不断舍去两侧的高度较低的线，使得面积更加集中。因此，我们可以通过双指针扫描法确定最大面积。
代码补充：
object Solution {
def maxArea(height: Array[Int]): Int = {
var maxArea = 0
var leftArr = 0
var rightArr = height.length-1
while (leftArr < rightArr) {
//println("-------------------------------------")
//println("leftArr: "+ leftArr)
//println("height(leftArr): "+height(leftArr))
//println("rightArr:" + rightArr)
//println("height(rightArr): "+height(rightArr))
//println("maxArea: "+maxArea)
var Area = (math.min(height(leftArr), height(rightArr)) * (rightArr-leftArr))
//if (Area < maxArea) return maxArea
//else {
if (height(leftArr) < height(rightArr)) leftArr+=1
else rightArr-=1
if (Area> maxArea)maxArea = Area
//}
}
return maxArea
}
}